Distance from vectors in $mathbb{Z}^d$ to the cube $[-1/2,1/2]^d$

Question

Let $|;|_2$ be the Eucliden norm on $mathbb{R}^d$.
Problem: Suppose $xin Q:=big[-tfrac12,tfrac12big]^d$. Is
$$
|x-k|_2geq frac{1}{2sqrt{d}}|k|_2
$$
for all $kinmathbb{Z}^d$?
Notice that for all $xin Q$,  $|x-k|_2geq d(k,Q):=inf_{xin Q}|x-k|_2$. So it is to enough to show that
$$d(k,Q)geq frac{|k|_2}{2sqrt{d}},quad kinmathbb{Z}^d$$

I think this holds but I can't reproduce a proof at the moment. The relevance of this simple geometric result is that it provides a criteria for absolute and uniform convergence of the Poisson summation
$$sum_{kinmathbb{Z}^d}f(x+k)=sum_{kinmathbb{Z}^d}widehat{f}(k)e^{2pi ikcdot x}$$
in the $mathbb{T}^d$ torus, where  $fin L_1(mathbb{R}^d)$. If $f$   can be bounded poitwise by an integrable decreasing radial function, i.e. $|f(x)|leq phi_0(|x|_2)$, where $phi_0$ is monotone non increasing and $phi_0circ|;|_2in L_1(mathbb{R}^d)$
A proof or a good hint will be appreciated.

Oliver Diaz · Accepted Answer

After giving it more thought, I came up with the following  solution below. I am still  interested  in accepting a more elegant solution which may make  clever use of the law of cosines and/or the triangle inequality without resorting to induction.

My solution:
As stated, $Q=big[-tfrac12,tfrac12big]^d$. For any $mathbf{k}=(k_1,ldots,k_d)inmathbb{Z}^d$, define
$$Delta(mathbf{k})={1leq jleq d: k_jneq0}$$
By compactness, there is $mathbf{q}inpartial Q$ such that
$$d(mathbf{q},mathbf{k})=d(mathbf{k},Q)$$
Since $0<big|m-operatorname{sign}(m)frac12big|leq |m-t|$ for all integer $mneq0$ and $|t|leqfrac12$
$$
d(mathbf{k},Q) =sum_{jinDelta(mathbf{k})}big(|k_j|-tfrac12big)^2
$$
That is, $mathbf{q}=(q_1,ldots,1_d)$ where $q_j=operatorname{sign}(k_j)frac12$ when $jinDelta(mathbf{k})$ and $q_j=0$ otherwise.
We proceed to show
$$begin{align}
d(mathbf{k},Q)geqfrac{|mathbf{k}|_2}{2sqrt{d}}tag{1}label{one}
end{align}
$$
or equivalently
$$begin{align}
4dsum_{jinDelta(mathbf{k})}(|k_j|-tfrac12)^2geqsum_{jinDelta(mathbf{k})}k^2_jtag{1'}label{onep}
end{align}$$
by induction on the dimension $D$ of the space.

For $D=1$, eqref{onep} holds trivially for $k=0$, and also for $kneq0$ since
$2big||k|-tfrac12big|=2|k|-1geq|k|$ since $|k|geq1$.

Asume eqref{onep} holds for $D=1,ldots,d-1$, where $d-1geq1$. Let $mathbf{k}inmathbb{Z}^d$ and $alpha_mathbf{k}=#Delta(mathbf{k})$.
If $alpha_mathbf{k}<d$, then
$$
begin{align}
4dsum_{jinDelta(mathbf{k})}(|k_j|-tfrac12)^2geq4alphasum_{jinDelta(mathbf{k})}(|k_j|-tfrac12)^2geq sum_{jinDelta(mathbf{k})}k^2_j
end{align}
$$
by the indiction hypothesis. If $alpha=d$, then $|k_j|>0$ for all $1leq jleq d$ and so,
begin{align}
4dsum_{jinDelta(mathbf{k})}(|k_j|-tfrac12)^2 &=4dsum^{d-1}_{j=1}(|k_j|-tfrac12)^2 + 4d(|k_d|-tfrac12)^2\
&geq  4(d-1)sum^{d-1}_{j=1}(|k_j|-tfrac12)^2 + 4(|k_d|-tfrac12)^2\
&geq sum^{d-1}_{j=1}k^2_j + k^2_d=|mathbf{k}|^2_2
end{align}

This completes the induction argument.

Distance from vectors in $mathbb{Z}^d$ to the cube $[-1/2,1/2]^d$

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